Under the stronger assumption REF  this is [1]}; let us describe how to weaken this assumption to REF  as outlined in [2]}. The idea is to iteratively construct quotients by the abelian groups \(U^{(j)}/U^{(j-1)}\)  for \(j =1 , \dots ,r\)  by locally choosing complementary subgroups to the stabiliser groups. To illustrate the argument, we assume for simplicity that \(U\)  is abelian; then by REF , there exists \(d\)  such that \(\dim \operatorname{Stab}_U(y) =d\)  for all \(y \in Y_{\min }\) . As \(U\)  is an abelian unipotent group, for each stabiliser group \(\operatorname{Stab}_U(y)\)  of \(y \in Y_{\min }\) , we can choose a complementary subgroup \(U^{\prime }\)  with \(\operatorname{Stab}_U(y)U^{\prime } = U\)  and \(\operatorname{Stab}_U(y)\cap U = \lbrace 1\rbrace \) , by choosing a complementary subspace in the Lie algebra. Taking the Lie algebra of the \(U\) -stabiliser gives a \(\widehat{U}\) -invariant function \(Y_{\min } \rightarrow \mathrm {Gr}(d,\operatorname{Lie}(U))\)  and being complementary to a given \(U^{\prime } < U\)  is an open condition in this Grassmannian. Therefore, we get a corresponding open \(\widehat{U}\) -invariant set of \(Y_{\min }\)  on which all stabilisers have complementary subgroup \(U^{\prime }\)  and thus quotienting by \(U\)  is equivalent to quotienting by \(U^{\prime }\) . This open set of \(Y_{\min }\)  admits a geometric \(U^{\prime }\) -quotient (and thus \(U\) -quotient) by [2]}. Then by varying \(U^{\prime }\)  we can cover \(Y_{\min }\)  by open sets admitting geometric \(U\) -quotients in order to obtain a geometric \(U\) -quotient of \(Y_{\min }\) . From there one proceeds as in the proof of the case when REF  holds.
